Is there some sort of monad whose algebras are monads? How about if we are internal to a bicategory B? Are internal monads in B monadic? Certainly not always, as otherwise free T-multicategories a la Leinster would always exist. What is known?

3 Answers
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In the book "Toposes, Triples and Theories", Barr and Wells study the question when a particular endofunctor admits a free monad. This is the case if the underlying category is complete and cocomplete and if the endofunctor preserves filtered colimits (we say that such a functor is finitary).

The question remains whether the resulting adjunction between monads on $C$ and endofunctors on $C$ is monadic. If $C$ is locally finitely presentable (lfp), this is true: Steve Lack showed in

that the forgetful functor $\mathrm{Mnd}_f(C) \rightarrow \mathrm{End}_f(C)$ from finitary monads on a lfp category $C$ to finitary endofunctors is monadic. Note that both these categories are again lfp categories, and we don't need universes to make sense of them, essentially because a finitary endofunctor is determined by what it does on a set of objects.

The result remains true if we consider categories and functors enriched in a complete and cocomplete symmetric monoidal closed category $V$ which is lfp as a closed category.

It's 83 pages long and famously impenetrable. But those who've braved it say that it contains just about every free monoid construction that anyone's ever thought of.

So if you want to construct a free monoid/monad, the question is whether it's easier to (a) read Kelly's paper, or (b) do it yourself and cite Kelly anyway (on the assumption that your construction is probably in there, somewhere).

In the book by Barr and Wells ("Toposes, Triples, Theories"), they describe the construction of a free monad on an endofunctor $F:C\to C$. They certainly require some conditions on $F$ in order to build the free monad, but I don't remember anymore what they are. I think that requiring $C$ to cocomplete and $F$ to commute with filtered colimits are sufficient conditions, but it's been a long time ...

This suggests that you can describe monads as algebras for the free monad monad, which will be a monad on a suitable full subcategory of the category $\mathrm{End}(C)$ of endofunctors of $C$, if $C$ is cocomplete.

Cool, thanks! This looks very reminiscent of the construction for the free T-multicategory monad when T is a finitary Cartesian monad. I bet this is not a coincidence.
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David CarchediApr 1 '10 at 13:47